Problem: What is the extraneous solution to these equations? $\dfrac{x^2 + 4}{x - 2} = \dfrac{12x - 16}{x - 2}$
Answer: Multiply both sides by $x - 2$ $ \dfrac{x^2 + 4}{x - 2} (x - 2) = \dfrac{12x - 16}{x - 2} (x - 2)$ $ x^2 + 4 = 12x - 16$ Subtract $12x - 16$ from both sides: $ x^2 + 4 - (12x - 16) = 12x - 16 - (12x - 16)$ $ x^2 + 4 - 12x + 16 = 0$ $ x^2 + 20 - 12x = 0$ Factor the expression: $ (x - 2)(x - 10) = 0$ Therefore $x = 2$ or $x = 10$ At $x = 2$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 2$, it is an extraneous solution.